# How to prove that there is no $\mathbb{E} : \mathbb{R}\subset\mathbb{E}$ which would have all those properties that $\mathbb{R}$ has?

That's a theorem from my calculus course: $\neg\exists\mathbb{E}\supset\mathbb{R}$ which would have all the $\mathbb{R}$ properties.

First part: $\neg\exists n > +\infty$ (that contradicts Archimedes axiom), same with $-\infty$.

But how about proving this: $\neg\exists e\in\mathbb{E} \hspace{3mm}e \notin \mathbb{R}:\exists x,y\in\mathbb{R}:x<e<y$.

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You should clarify what you mean by 'properties'. For example, if such a set $\mathbb{E}$ exists and properly contains $\mathbb{R}$, then it has an element which isn't contained in $\mathbb{R}$, which means that $\mathbb{E}$ doesn't have the property of not containing any non-real elements. Does this count? –  Clive Newstead Jan 15 '12 at 15:11
You really need to make your question self-contained. What are "the $\mathbb R$ properties." –  Thomas Andrews Jan 15 '12 at 15:11
@Henning: If you make that an answer, I’ll upvote it. –  Brian M. Scott Jan 30 '12 at 7:01
@Henning: That’s too bad; it would have disposed of the question pretty thoroughly. –  Brian M. Scott Jan 30 '12 at 21:33

You may be looking for something like Theorem 15 here.

(converted from a comment, as suggested by Brian M. Scott)

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